Chapter 12. Properties of Solutions
What we will learn: • Types of solutions • Solution process • Interactions in solution • Types of concentration • Concentration units • Solubility and temperature • Solubility and pressure • Colloids
GCh12-1 Solution
• A homogeneous mixture of two or more substances
Saturated solution Unsaturated solution
• Contains the maximum amount • Contains less amount of a solute of a solute that will dissolve in a that will dissolve in a solvent solvent
Supersaturated solution Crystalization
• Contains more amount of a solute • The process in which a dissolved that will dissolve in a solvent (an solute comes out of solution and unstabile state) forms cristals
GCh12-2 Types of solutions
Comp. 1 Comp. 2 State Examples Gas Gas Gas Air Gas Liquid Liquid Soda water
Gas Solid Solid H2 in palladium Liquid Liquid Liquid Ethanol in water Solid Liquid Liquid NaCl in water Solid Solid Solid Brass (Cu/Zn)
GCh12-3 Molecular view of solution process
During the solution process, solvent molecules interact with solute molecules. If the solvent-solute interactions are stronger than the solvent-solvent and solute-solute interactions, then the solute is dissolved in the solvent. (Energy driven process)
Another driving force of the solution process is an increase of dissorder on the final system. If the dissorder of the solution is bigger then dissorder of the solvent and dissorder of the solute, then the solute is dissolved in the solvent. (Entropy driven process)
Three steps of solution process (energy driven)
D • Heat of separation of solute molecules H1 D • Heat of separation of solvent molecules H2 D • Heat of mixing of solute and solvent molecules H3
GCh12-4 Heat of solution
D D D D Hsol = H1 + H2 + H3
Example
The endothermic process - solution Exothermic of sodium chloride in water (a favorable process)
D Hsol < 0
Endothermic (a non favorable process)
D Hsol > 0
GCh12-5 Solubility
• Measure of how much a solute will dissolve in a solvent at a particular temperature
Like dissolves like
• A solute is better dissolved in a solvent, which has similar intermolecular force types and magnites to a solute
Ionic molecules are better dissolved Examples in polar solvents
• H2O is a good solvent for NaCl Nonpolar molecules are better dissolved in nonpolar solvents • CCl4 is a good solvent for C6H6
GCh12-6 Hydrogen bondings between water and methanol are responsible for solubility of methanol in water
Water molecules are stabilizing Na+ and Cl- ions in NaCl solution
Dispersion interactions between benzene
and CCl4 are responsible for solubility of
CCl4 in benzene
GCh12-7 Miscible liquids
• Two liquids which are completely soluble in each other in all proportions
Example
• Methanol and water
• Ethanol and water
• 1,2-ethylene glycol and water
GCh12-8 Solvation
• The process in which a solute molecule (ion) is surrounded by solvent molecules
There are several solvation spheres. The solvation process involving water molecules is called hydration
GCh12-9 Problem
Predict the relative solubility in the following cases:
a) Bromine (Br2) in benzene (C6H6) and in water b) KCl in carbon tetrachoride (CCl4) and in liquid ammonia (NH3) c) Formaldehyde (CH2O) in carbon disulfide (CS2) and in water
m = 1.87 D m = 2.33 D m = 1.46 D
GCh12-10 Concentration units
• Percent by mass (units: %), is independent of temperature and does not require calculations of molar masses
• Mole fraction (no units); is used in calculating partial pressures of gases
• Molarity (units: mol/L); used practically in laboratories because a volume of a solution is easy to measure, but is dependent of temperature
• Molality (units: mol/kg); is independent of temperature because the concentration is expressed in solute number of moles and mass of solvent
GCh12-11 Percent by mass
mass of solute (msl) = X 100% (units %)
mass of solute (msl) + mass of solvent (msv)
Example
What is the percent by mass of 0.892 g KCl dissolved in 54.6 g water.
Data
msl - mass of solute = 0.892 g msv - mass of solvent = 54.6 g
Solution percent by mass = 0.892 g / (0.892 g + 54.6 g) X 100 %
= 1.61 %
GCh12-12 Mole fraction
moles of solute (nsl) X = (no units)
moles of solute (nsl) + moles of solvent (nsv)
Example
What is the mole fraction of 12.5 g sodium chloride (NaCl) dissolved in 80.5 g water.
Data Number of water moles: m - mass of solute = 12.5 g sl 1 mol = 18.02 g msv - mass of solvent = 80.5 g n mol = 80.5 g Solution n = 1 mol x 80.5 g / 18.02 g Step 1. Calculation of moles: = 4.46 mol
nsv = 4.46 mol GCh12-13 GCh12-14 Step 2. Calculation of moles Number of NaCl moles:
nsl = 0.21 mol 1 mol = 58.44 g
n mol = 12.5 g
n = 1 mol x 12.5 g / 58.44 g = 0.21 mol Step 3. Calculation of mole fraction mole fraction = 0.21 mol / (0.21 mol + 4.46 mol)
= 0.04
GCh12-15 Molarity
moles of solute (nsl) M = (units: mol / L)
volume of solution (Vsn)
Example
Calculate molarity of a 1.2 L of HNO3 solution containing 24.4 g of HNO3.
Data
msl - mass of solute = 24.4 g
Vsn - volume of solution = 1.2 L
GCh12-16 Solution
Number of HNO3 moles Step 1. Calculation of moles 1 mol = 63.02 g
nsl = 0.38 mol n mol = 24.4 g
Step 2. Calculation of molarity n = 1 mol x 24.4 g / 63.02 g = 0.38 mol molarity = 0.38 mol / 1.2 L
= 0.31 mol/L
GCh12-17 Molality
moles of solute (nsl) m = (units: mol / kg)
mass of solvent (msv)
Example
Calculate the molality of H2SO4 solution, containing 24.4 g of H2SO4 in 198 g of water.
Data
msl - mass of solute = 24.4 g msv - mass of solvent = 198 g = 0.98 kg
GCh12-18 Solution Number of H SO moles Step 1. Calculation of moles: 2 4 1 mol = 98.10 g nsl = 0.24 mol n mol = 24.4 g
n = 1 mol x 24.4 g / 98.10 g Step 2. Calculation of molality: = 0.24 mol
nsl = 0.24 mol /0.98 kg
= 1.26 mol / kg
GCh12-19 Density of solution
mass of solution (msn) d = (units: g / mL)
volume of solution (Vsn)
Examples
msn - mass of solution = 128 g
Vsn - volume of solution = 1.45 L = 1450 mL
d = 128 g / 1450 mL = 0.08 g/mL
GCh12-20 Problem
The molarity if a aqueous solution of methanol is 2.45 M. The density of this solution is 0.976 g/mL. What is the molality of the soltuion.
Data
dsn - density of solution = 0.976 g/mL
Msn - molarity of solution = 2.45 M
Solution
moles of solute (nsl) mass of solution (msn) M = d =
volume of solution (Vsn) volume of solution (Vsn)
moles of solute (nsl) m =
mass of solvent (msv)
GCh12-21 Step 1. Calculation of mass of solute Mass of CH OH: 2.45 moles - 1 liter of solution 3 1 mol = 32.04 g 79.49 g - 1 liter of solution 2.45 mol = n g
n = 2.45 mol x 32.04 g / 1mol = 79.49 g Step 2. Calculation of mass of water
0.976g of solution - 1 mliter of solution
976 g of solution - 1 liter of solution mass of water = 976 g - 76.49 g = 898 g
GCh12-22 Step 3. Calculation of molality: m = 2.45 mol / 0.898 kg
= 2.73 m
GCh12-23 Problem
The density of an aqueous solution containing 10 % of ethanol (C2H2OH) by mass, is 0.984 g/mL. Calculate the molality of this solution. Calculate the molarity
Data
dsn - density of solution = 0.984 g/mL P - percent by mass of solution = 10%
moles of solute (n ) sl mass of solution (msn) M = d = volume of solution (V ) sn volume of solution (Vsn)
moles of solute (nsl) m =
mass of solvent (msv)
GCh12-24 Step 1. Calculation of mass of water in 100 g Moles of ethanol: of solution 1 mol = 46.07 g 10g (E) - 100g (S) n mol = 10 g m (W) = 100 g (S) - 10 g(E) n = 10 g x 1 mol / 46.07 g = 90 g = 0.09 kg = 0.217 mol
Step 2. Calculation of ethanol moles in 100 g of solution
10g (E) (1 mol) / (46.07 g) = 0.217 mol (E)
Step 3. Calculation of molality of 100 g of solution m = (0.217 mol) / (0.09 kg)
= 2.41 mol/kg
GCh12-25 Step 4. Calculation of a volume of 100 g of Volume of solution: solution 1 mL = 0.984 g 100 g (1 mL ) / (0.984 g) = 102 mL = 0.102 L V mL = 100 g
V = 100 g x 1 mL / 0.984 g = 102 mL
Step 5. Calculation molarity of 100 g of solution
M = (0.217 mol) / (0.102 L) = 2.13 mol/L
GCh12-26 Effect of temperature on solubility Solubility • Solubility of solid solutes • Maximal concentration of a • Solubility of gas solutes solute in a solvent
Solid solubility
Generally, the solubility of solids is bigger at the higher temperature
There is no correlation between the solubility of a solute at a particular temperature, and its heat of solution D Hsol
• CaCl2 exothermic
• NH4NO3 endothermic
Solubility of both compounds increases with the temperature GCh12-27 Fractional crystalization
• Separation of a solution into pure solutes using their different solubilities
By lowering temperature we can extract from a solution a pure solute.
If we have a solution of a mixture of two solutes, by using fractional crystalization we can separate those two solutes.
GCh12-28 Gas solubility
Generally, the solubility of gasses is smaller at the higher temperatures
Fish prefer to be deeper in water, where the temperature is lower, and the oxygen concentration is bigger
Global warming decreses the concentration of oxygen in water, which is harmful to its living animals
GCh12-29 Effect of pressure on gas solubility
Henry’s law
• The solubility of a gas in a liquid is proportional to the pressure of the gas over the liquid
c = kP c - gas concentration (molarity) k - constant P - pressure (partial)
Dynamic equilibration
• A number of molecules going from the gas to the liquid is the same as the number of molecules going from the liquid to the gas at the same time GCh12-30 Practically, in soft drinks, the pressure of CO2 in a bottle is big which makes bigger the concentration CO2 in the liquid
Problem
The solubility of pure nitrogen gas in water at 25 C and at 1 atm, is 6.8 x 10-4 mol/L. What is the concentration (in molarity) of nitrogen dissolved in water under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm.
Data -4 cN2 - solubility of nitrogen gas = 6.8 x 10 mol/L
PN2 - partial pressure of nitrogen gas = 0.78 atm P - atmospheric pressure = 1 atm
c = kP k = c/P
k = 6.8 x 10-4 mol/L atm
GCh12-31 c = ( 6.8 x 10-4 mol/L atm ) ( 0.78 atm )
= 5.3 x 10-4 mol/L
Colligative (collective) properties of nonelectrolyte solution
• Properties which depend on the number of solute particles in solution and not on the nature of solute
GCh12-32 Vapor-pressure lowering
• The vapor pressure of its solution is always smaller than that of the pure solvent
For the solute, which is nonvolatile (it does not have a vapor pressure)
Raoult’s law
0 P1 = X1P1
P1 - vapor pressure of a solvent over a solution 0 P1 - vapor pressure of the pure solvent
X1 - mole fraction of the solvent
X1 = 1 - X2
X2 - mole fraction of the solute
GCh12-33 Problem
Calculate the vapor pressure of a solution containing 218 g glucose in 460 mL of water Moles of water: at 30 C. The vapor pressure of pure water at 30 C is 31.82 mmHg. The density of the 1 mol = 18.02 g solution is 1 g/mL.
n1 mol = 460 g Data 0 n = 1 mol x 460 g/18.02 g P1 - vapor pressure of pure water = 31.82 1 mmHg = 25.5 mol m2 - mass of the solute = 218 g Moles of glucose: V1 - volume of water = 460 mL d1 - density of water = 1 g / mL 1 mol = 180.2 g m1 - mass of water = 460 g
Step 1. Calculate the number of moles n2 mol = 218 g
n = 1 mol x 218 g/180.2 g n1 = 25.5 mol 2 = 1.21 mol n2 = 1.21 mol
GCh12-34 Step 2. Calculation of the water mole fraction
X1 = 25.5 mol / (25.5 mol + 1.21 mol)
X1 = 0.955
Step 3. Calculation of the vapor pressure
0 P1 = X1P1
P1 = 0.955 x 31.82 mmHg
= 30.4 mmHg
GCh12-35 Raoult’s law (a more practical equation)
D 0 P = X2P1
DP - vapor pressure lowering over a solution 0 P1 - vapor pressure of the pure solvent
X2 - mole fraction of the solute
Problem
At 25 C the vapor pressure of pure water is 23.76 mmHg and that of seawater is 22.98 mmHg. Assuming that seawater contains only NaCl, estimate its molal concentration (molality).
GCh12-36 Data
D 0 P = X2P1
0 P1 - vapor pressure of pure water = 23.76 mmHg
P1 - vapor pressure of solution = 22.98 mmHg
D 0 P = X2P1
D 0 X2 = P / P1
= (23.76 mmHg - 22.98 mmHg) / (23.76 mmHg)
= 0.033 assuming that we have 1000 g of water as the solvent
n1 = (1000 g ) / (18.02 g ) = 55.49 mol
X2 = n2 / (n1 + n2)
GCh12-37 n2 / (55.49 + n2) = 0.033
n2 = 1.88 mol
Since NaCl dissociates to form two ions, the number of moles of NaCl is half of the above results
Moles NaCl = 0.94 mol
The molality of the solution
m = (0.94 mol) / (1 kg) = 0.94 mol/kg
GCh12-38 Vapor pressure over a solution of two liquids
Two volatile liquids (both Raoult’s law having a vapor pressure)
0 PA = XAPA
0 PB = XBPB Example
PA - vapor pressure of a liquid A over a solution Benzene and toluene 0 PA - vapor pressure of the pure liquid A PB - vapor pressure of a liquid B over a solution A - benzene 0 PB - vapor pressure of the pure liquid B XA - mole fraction of the liquid A B - toluene XB - mole fraction of the liquid B
PT = PA + PB
PT - total vapor pressure over the solution
GCh12-39 Case 1. DH of solution is zero (ideal solution)
The forces between A and B molecules are the same as the forces between A and A, and between B and B.
No termal effects of the solution process
The Raoult’s law is fulfilled
Example
A mixture of benzene and toluene
GCh12-40 Case 2. DH of solution bigger than zero (endothermic process)
The forces between A and B molecules are smaller than the forces between A and A, and between B and B. Molecules of both solutes have a tendency to leave the solution and go to the gas phase. The total vapor pressure is bigger than the sum of partial vapor pressures.
The solution process requires the energy
The positive deviation of the Raoult’s law
Example:
The system of chloroform (CHCl3) and ethanol (CH3CH2OH)
GCh12-41 Case 3. DH of solution smaller than zero (exothermic process)
The forces between A and B molecules are bigger than the forces between A and A, and between B and B. Molecules of both solutes have a tendency to be in the solution rather than go to the gas phase. The total vapor pressure is smaller than the sum of partial vapor pressures.
The solution process generates the energy
The negative deviation of the Raoult’s law
Example:
The system of chloroform (CHCl3) and acetone (CH3COCH3)
GCh12-42 Fractional distillation
• Separating a solution of two liquids into individiual components, based on their different boiling points.
Practically used in chemical laboratories and industry (oil fractional distillation in petroleum refineries)
GCh12-43 Boiling point
• The temperature at which the vapour pressure of a solution is equal to the external amospheric pressure
A solid solute usually lowers the vapor pressure of its solution, therefore the boiling point of the solution is bigger than the boiling point of the pure solvent
Boiling point elevation
• This is a difference between the boiling point of the solution and the boiling point of the pure solvent
GCh12-44 D 0 Tb = Tb - Tb
Tb - boiling point of the solution 0 Tb - boiling point of the pure solvent
The boiling point of the solution depends on the solution concentration (molality)
D Tb = Kb m
Kb - molar boiling-point elevation constant m - molality concentration
GCh12-45 Freezing point
• The temperature at which a liquid solution transforms into a solid
When the solution freezes, its separates into individual solvent components
Freezing point depression
• This is a difference between the freezing point of the solution and the freezing point of the pure solvent
GCh12-46 D 0 Tf = Tf - Tf
Tf - freezing point of the solution 0 Tf - freezing point of the pure solvent
The freezing point of the solution depends on the solution concentration (molality)
D Tf = Kf m
Kf - molar freezing-point depression constant m - molality concentration
GCh12-47 Problem
Calculate the freezing point of a solution containing 651 g of ethylene glycol in 2505 g of water. The molal freezing-point depression constant of water is 1.86 C/m.
Data
msu - mass of solute = 651 g msl - mass of solvent = 2505 g
Kf - molal freezing-point depression constant of solvent = 1.86 C/m
Solution Moles of ethylene glycol: DT = K m f f 1 mol = 62.01 g
n mol = 651 g
n = 651 g x 1 mol / 62.01 g = 10.5 mol
GCh12-48 m = (moles of solute) / (mass of solvent)
m = (10.5 mol) / (2.505 kg)
m = 4.19 m
D Tf = (1.86 C/m) (4.19 m)
= 7.79 C
GCh12-49 Osmosis
• An effect of moving solvent molecules through a membrane from a less to a more concentrated solution
Solvent molecules are moving from a smaller dissorder state in a less concentrated solution, to a bigger dissorder state in a more concentrated solution
Entropy driven process
GCh12-50 Osmotic pressure
• A pressure which stops the osmotic process
p = MRT p - osmotic pressure M - molarity of the solution R - constant (gas constant = 0.0821 L atm/K mol) T - temperature
GCh12-51 The osmotic pressure is responsible for moving water from tree roots to tree leafs
Problem
The osmotic pressure between two solutions of seawater is 30.0 atm. Calculate the molar concentration of sucrose that is isotonic with seawater. (Isotonic: two solutions having the same concentration, and therefore the same osmotic pressure)
Data p - osmotic pressure = 30.0 atm T - temperature = 25 C = 298 K R - gas constant = 0.0821 (L atm)/(K mol)
GCh12-52 p = MRT
M = p /(RT)
= 30.0 atm / [ (0.0821 L atm/K mol) (298 K) ]
= 1.23 mol/L
GCh12-53 Determination of molar mass
The freezing-point depression or osmotic pressure, we can determine the molar mass of a solvent in a solution
Problem
A 7.85 g of an unknown compound is dissolved in 301 g of benzene. The freezing point of the solution is 1.05 C below that of pure benzene. What is molar mass of the compound ?
Data
msu - mass of a solute = 7.85 g msl - mass of a solvent = 301 g D Tf - freezing-point depression = 1.05 C
Kf - freezing-point depression constant of benzene = 5.12 C/m
GCh12-54 Step 1 Number of moles:
D Tf = Kf m 0.205 mol = 1 kg
D m = Tf / Kf n mol = 0.301 kg
= 1.05 C/5.12 (C/m) n = 0.301 kg / 0.205 mol
= 0.205 m = 0.0617 mol
Step 2 molar mass = (grams of compund) / (moles of compound)
= 7.85 g / 0.0617 mol
= 127 g/mol
GCh12-55 Problem
A solution is prepared by dissolving 35 g of hemoglobin (Hb) in enough water to make up 1 L in volume. The osmotic pressure of the solution is 10 mmHg at 25 C. What is the molar mass of hemoglobin.
Data
msu - mass of the solute = 35 g
Vsn - volume of the solution = 1 L p - osmotic pressure = 10 mmHg = 0.0132 atm T - temperature 25 C = 298 K R - gas constant = 0.0821 (L atm)/(K mol)
p = MRT
M = p /(RT)
= ( 0.0132 atm ) / [(0.0821 L atm / K mol) (298 K)]
= 5.38 x 10-4 M
GCh12-56 The volume of the solution is 1 L, therefore
1 L (solution) g 5.38 x 10-4 mol Hb moles of Hb = (mass of Hb) / (molar mass of Hb) molar mass of Hb = (mass of Hb) / ( moles of Hb )
= (35 g ) / (5.38 x 10-4 mol)
= 6.51 x 104 g / mol
GCh12-57 Colligative (collective) properties of electrolyte solution
• Properties which depend on the number of solute particles after solute dissociation in solution, and not on the nature of solute
van’t Hoff factor
actual number of particles after dissociation i = number of formula units initially dissolved
For most non-electrolytes dissolved in water, the van’ t Hoff factor is essentially 1 Example For most ionic compounds dissolved in water, NaCl i = 2 the van ‘t Hoff factor is equal to the number Na SO i = 3 of discrete ions in a formula unit of the 2 4 substance.
GCh12-58 D Tf = i Kf m
D Tb = i Kb m
p = i MRT
Problem
The osmotic pressure of a 0.010 m potassium iodide (KI) solution at 25 C is 0.465 atm. Calculate the van’t Hoff factor for KI at this concentration?
Data p - osmotic pressure = 0.465 atm M - molarity of the solution = 0.010 m T - temperature = 25 C = 298 K R - gas constant = 0.082 L atm/K mol
GCh12-59 Solution
p = i MRT
i = p/MRT
= 0.465 atm/ [ (0.010 m) (0.0821 Latm/Kmol) (298K) ]
= 1.90
GCh12-60 The theoretical van’t Hoff’s factor is sometimes different from the experimentally observed
Electrolyte i(measured) i(calculated) HCl 1.9 2 NaCl 1.9 2
MgSO4 1.3 2
MgCl2 2.7 3
FeCl3 3.4 4
Ionic pair
Ions of opposite signs which are held together be electrostatic forces
Electrostatic interactions between ions form ionic pairs. The presence of ionc pairs reduces the effective number of particles in solution
GCh12-61 Colloids
• A dispersion of particles of one substance through a dispersing medium of another substance
Water colloids
Hydrophilic (water-loving)
Hydrophobic (water-fearing)
GCh12-62 How soap works
Soap has a hydrophilic head, and a hydrophobic tail. Hydrophobic tails of soap molecules interact with a dust particle, making the dust particle soluble in water
GCh12-63